Slope of curve in $\mathbb{R}^3$

While doing revision, I came across this problem:

The surface given by $z=x^2-y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane. Find the slope of this curve at the point $(1,3,-8)$.

I tried substituting $y=3x$ into $z=x^2-y^2$, yielding $z=-8x^2$. Then, $\frac{dz}{dx}=-16x=-16$.

However the answer is $-8\sqrt{\frac{2}{5}}$.

Thank you very much for any help.

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What do you mean by slope of a line in dimension 3? –  Mercy Jun 23 '12 at 11:02
@Mercy I think the question requires us to consider the shape of the intersection in the plane $y=3x$. Re orientation I suppose we should regard $z$ as the dependent variable. –  hwhm Jun 23 '12 at 11:14
According to mathworld.wolfram.com/Slope.html; It is meaningless to talk about the slope of a curve in three-dimensional space unless the slope with respect to what is specified. –  Babak S. Jun 23 '12 at 11:18
This curve can be parametrized by $\gamma(t)=(t,3t,-8t^2)$ so I don't really understand the notion of slope here! –  Mercy Jun 23 '12 at 11:21
@Mercy You might wanna do a 3D plot. The question is badly posed but it's clear from the plot that what's wanted is to have axes introduced to the plane $y=3x$ (I'd say the horizontal axes is along $(1,3,0)$ and the vertical along $(0,0,1)$) and then talk about the intersection as though it is graphed on that plane. –  hwhm Jun 23 '12 at 11:25

The following is a solution to a rephrased question which can be answered, however.

Question: The surface given by $z=x^2−y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane.

Treating the intersection as a curve in the said plane with vertical axis along $(0,0,1)$ and horizontal axis along $(1,3,0)/\sqrt{10}$, find the slope of this curve at the point (1,3,−8).

Solution: Any point on the plane has Cartesian coordinates in the form $$\frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix}.$$ Substituting this into $z=x^2−y^2$, we get $b = -4a^2/5$.

So the "slope" at a point on this intersection, with $a$ and $b$ given, is $$\frac{db}{da} = -\frac{8a}{5}.$$

Setting $$\begin{pmatrix}1 \\3\\-8\end{pmatrix} = \frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix},$$ we get $a = \sqrt{10}$ and so the "slope" at this point is $-8\sqrt{\frac{2}{5}}$.

Solution using grad: Let $f:=x^2-y^2-z$ and $g:=y-3x$. At the point $(1,3,-8)$, $\nabla f=(2,-6,-1)$ and $\nabla g=(-3,1,0)$. Their cross product, $(1,3,-16)$, is along the tangent direction of the intersecting curve produced by the surface and the plane, at the point $(1,3,-8)$.

Denote the angle between $(1,3,-16)$ and $(1,3,0)$ (i.e. the "horizontal") by $\theta$. Then, using the dot product, $\cos\theta = \sqrt{\frac{5}{133}}$. The "slope" is $$\tan \theta = - \sqrt{\frac{1}{\cos^2 \theta}-1} = -8\sqrt{\frac{2}{5}},$$ where the negative square root is taken because the "vertical" is along $(0,0,1)$.

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Thanks a lot. Is there any way to solve this question using the Gradient operator? ($\nabla F$) –  yoyostein Jun 24 '12 at 7:13
@yoyostein Of course - see the updated answer. I didn't know the question was meant to be undergrad level then - it did look like a very badly phrased high school level question! –  hwhm Jun 25 '12 at 3:55